Question

Graph △ABC after a rotation of 270° clockwise about the origin.
A'=
B'=
C'=

Answer 1

so what is the answer

Answer 2

To rotate a point 270° clockwise about the origin, we can use the following rotation matrix:

\[
\begin{bmatrix}
\cos(270°) & -\sin(270°) \\
\sin(270°) & \cos(270°)
\end{bmatrix}
\begin{bmatrix}
x \\
y
\end{bmatrix}
\]

Since we are given the coordinates of points A, B, and C, we can substitute these values into the rotation matrix to find the new coordinates after the rotation.

Let's assume the coordinates of point A are (x1, y1), of point B are (x2, y2), and of point C are (x3, y3). After substituting these values into the rotation matrix, we get:

A' =
\[
\begin{bmatrix}
\cos(270°) & -\sin(270°) \\
\sin(270°) & \cos(270°)
\end{bmatrix}
\begin{bmatrix}
x1 \\
y1
\end{bmatrix}
\]

B' =
\[
\begin{bmatrix}
\cos(270°) & -\sin(270°) \\
\sin(270°) & \cos(270°)
\end{bmatrix}
\begin{bmatrix}
x2 \\
y2
\end{bmatrix}
\]

C' =
\[
\begin{bmatrix}
\cos(270°) & -\sin(270°) \\
\sin(270°) & \cos(270°)
\end{bmatrix}
\begin{bmatrix}
x3 \\
y3
\end{bmatrix}
\]

Now, let's calculate the values using the rotation matrix:

A' =
\[
\begin{bmatrix}
\cos(270°) & -\sin(270°) \\
\sin(270°) & \cos(270°)
\end{bmatrix}
\begin{bmatrix}
x1 \\
y1
\end{bmatrix}
\]

A' =
\[
\begin{bmatrix}
0 & -1 \\
1 & 0
\end{bmatrix}
\begin{bmatrix}
x1 \\
y1
\end{bmatrix}
\]

A' =
\[
\begin{bmatrix}
0 \cdot x1 + -1 \cdot y1 \\
1 \cdot x1 + 0 \cdot y1
\end{bmatrix}
\]

A' =
\[
\begin{bmatrix}
-y1 \\
x1
\end{bmatrix}
\]

Similarly, we can calculate the values for B' and C':

B' =
\[
\begin{bmatrix}
-y2 \\
x2
\end{bmatrix}
\]

C' =
\[
\begin{bmatrix}
-y3 \\
x3
\end{bmatrix}
\]

Answer 3

The answer is:

A' = (-y1, x1)
B' = (-y2, x2)
C' = (-y3, x3)